googologywikiaorg-20200223-history
User blog:Hyp cos/Analysis - BEAF, FGH and SGH (part 1)
Do you think the first time SGH catches up FGH at LVO or \(\psi(\Omega_\omega)\) ? (See here.) Do you think the limit of a legion of BEAF is LVO now? We used to think the limit of a legion of BEAF is LVO, which "happens to be" the first catching ordinal some people think. Now we know the real catching ordinal, so let's analysis BEAF again. I hope to see the real strength of BEAF. Maybe it's stronger than BAN. Let's go. Linear arrays In SGH, the n in \(g_\alpha(n)\) never change because \(g_{\alpha+1}(n)=g_\alpha(n)+1\) and \(g_\alpha(n)=g_{\alphan}(n)\) if a limit ordinal. Then the ω's in SGH mean always n and they can change to n freely. And ω's are "orderless" in SGH. In FGH and HH we cannot do that because the n is changing. When SGH grows a part of its ordinal from 2,3,4,... to ω (it's new, called "active point"), and at the same time FGH grows its ordinal to a limit one, then FGH ordinal increasing by 1 will change the SGH active point (the new ω) into Ω in ordinal collasping functions. From dimensional to tetrational arrays Question: What's the active point of θ(Ω^Ω) ? I can't find any ω's. Answer: An Ω in ordinal collasping function means ω nests, or n nests in SGH. Then θ(Ω^θ(Ω^Ω),θ(Ω^Ω)+1) means ω+1 nests (or n+1 nests in SGH), and θ(Ω^Ω,1) means ω2 nests (or 2n nests in SGH). Amazingly, what an "&n" does is --- Mapping SGH into FGH of the same ordinal approximately! That means, if an array function p(n) has growth rate α in SGH, p(X)&n will have growth rate α in FGH approximately (more accurate: ω^α in FGH). And what a normal array does to SGH will be what an (X-array)&n does to FGH. Linear array of arrays Now you see, we get LVO in FGH here, not a legion. In Bowers's page he uses things like "b&b&...b&b - p times", but the & operator has 2 properties:- One is "sequence" - the & strings don't output a number, and they catch this structure into the outside arrays. e.g. {3&3(1)3&3}={3,3,3(1)3,3,3}, and it's not {tritri(1)tritri}=tritri. The other is "holdleft" - things lie left of & can't be solve before we solve the &. e.g. triakulus=3&3&3 but it's not tritri&3={3,tritri(1)2}={3,3,2(1)2}. The second property make it hard to solve this arrays, so we have to use some symbols. Let X represent a line of n's in normal arrays, X2 represent a line of X's in X-arrays, and Xk+1 represent a line of Xk's in Xk-arrays. Now "b&b&...b&b - p times" become \(X_{p-1}\&X_{p-2}\&...X_2\&X\&b\). You see, the & string is "layer-by-layer", not linear. Don't misunderstand it to X&X&...X&b ! So {X,n+1(1)2}&n is approximately {X,X(1)2}&(n+1), which is much less than {X,X+1(1)2}&n. And X2+1&X&n is far above them because it's {X,X,...X(1)X}&n - X X's in the first row. By the way, in "{X,X,1,...1,2} 2 a.p X+2", position X+2 doesn't mean "the 2nd entry in the 2nd row in the X-array" - it's just the X+2-th entry in the first row in the X-array. Notice that X only means a line of n in normal arrays, not a line of everything including X in X-arrays. To represent a line of X, we should use X2. Also, using Bowers's original notation and the "clear" notation, triakulus=3&3&3=X2&X&3, golapulus=10^100&10&10=X2^100&X&10, golapulusplex=10^100&10&10&10=X3^100&X2&X&10. Does it make sense now? Okey. Next I'll continue the comparisons. Part 2 Category:Blog posts